Asked by Sjk
A student used 20 ml of 10 % potassium iodide rather than 15 ml of 10 % potassium iodide, what is the effect of this increase in potassium iodide on the calculated % NaOCl in the bleach?
Answers
Answered by
DrBob222
I am assuming that you are treating NaOCl with KI solution and the NaOCl oxidizes the KI to I2 and the I2 is titrated with a standard solution of Na2S2O3.
OCl^- + 2I^- + 2H^+ ==> I2 + Cl^- + H2O
I2 + 2S2O3^- ==> 2I^- + S4O6^2-
The limiting reagent here is the NaOCl. That's what you're determining. You add an excess of KI solution in the first place so a little extra excess won't make any difference. The amount of I2 liberated is a function of what? It's a function, not of the amount of KI present but a function of the amount of NaOCl present. That's what determines how much I2 is formed and the i2 is what's being titrated. Hope this helps.
OCl^- + 2I^- + 2H^+ ==> I2 + Cl^- + H2O
I2 + 2S2O3^- ==> 2I^- + S4O6^2-
The limiting reagent here is the NaOCl. That's what you're determining. You add an excess of KI solution in the first place so a little extra excess won't make any difference. The amount of I2 liberated is a function of what? It's a function, not of the amount of KI present but a function of the amount of NaOCl present. That's what determines how much I2 is formed and the i2 is what's being titrated. Hope this helps.
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